Physics-
General
Easy

Question

The efficiency of a Carnot engine is 50% and temperature of sink is 500 K. If temperature of source is kept constant and its efficiency raised to 60%, then the required temperature of sink will be

  1. 100 K    
  2. 600 K    
  3. 400 K    
  4. 500 K    

The correct answer is: 400 K


    As eta equals 1 equals fraction numerator T subscript 2 end subscript over denominator T subscript 1 end subscript end fraction
    therefore blank fraction numerator 50 over denominator 100 end fraction equals 1 equals fraction numerator 500 over denominator T subscript 1 end subscript end fraction o r T subscript 1 end subscript equals 1000 K
    Again, fraction numerator 60 over denominator 100 end fraction blank equals 1 minus fraction numerator T subscript 2 end subscript over denominator 1000 end fraction
    Or T subscript 2 end subscript equals 400 blank K

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